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Given an unknown filter B(Z), to understand its inverse,
we need to
factor B(Z) into two parts:
B(Z) = B_{out}(Z) B_{in}(Z), where B_{out} contains all the
roots outside the
unit circle and B_{in} contains all the roots inside.
Then the inverse of B_{out}
is expressed as a Taylor series about the origin,
and the inverse of B_{in} is expressed as a Taylor series about infinity.
The final expression for 1/B(Z)
is called a ``Laurent expansion'' for 1/B(Z),
and it converges on a ring including the unit circle.
Cases with zeros exactly on the unit circle present special problems.
For example, the differentiation filter (1Z)
is the inverse of integration,
but the converse is not true,
because of the additive constant of integration.
EXERCISES:

Find the filter that is inverse to (2  5Z + 2Z^{2}).
You may just
drop higherorder powers of Z, but an exact expression for the
coefficients of any power of Z is preferable.
(Partial fractions is a useful, though not a necessary, technique.)
Sketch the impulse response.

Describe a general method for determining A(Z) and B(Z) from a
Taylor series of
,where B(Z) and A(Z) are polynomials of unknown degree n and m,
respectively.
Work out the case
.Do not try this problem unless you are
familiar with determinants.
(HINT: identify coefficients of B(Z) = A(Z) C(Z).)
Next: INTRODUCTION TO ALLPASS FILTERS
Up: MINIMUMPHASE FILTERS
Previous: Mechanical interpretation
Stanford Exploration Project
10/21/1998